Let $R$ be a Riemann surface, $\widetilde{R}$a topological surface, and $p: \widetilde{R} \rightarrow R$ a continuous map. Suppose that every point $x \in \widetilde{R}$admits a neighbourhood $\widetilde{U}$such that $p$ maps $\widetilde{U}$homeomorphically onto its image. Prove that $\widetilde{R}$has a complex structure such that $p$ is a holomorphic map.

A holomorphic map $\pi: Y \rightarrow X$ between Riemann surfaces is called a covering map if every $x \in X$ has a neighbourhood $V$ with $\pi^{-1}(V)$ a disjoint union of open sets $W_{k}$ in $Y$, so that $\pi: W_{k} \rightarrow V$ is biholomorphic for each $W_{k}$. Suppose that a Riemann surface $Y$ admits a holomorphic covering map from the unit $\operatorname{disc}\{z \in \mathbb{C}:|z|<1\}$. Prove that any holomorphic map $\mathbb{C} \rightarrow Y$ is constant.

[You may assume any form of the monodromy theorem and basic results about the lifts of paths, provided that these are accurately stated.]